【codeforces 314C】Sereja and Subsequences

【题目链接】:http://codeforces.com/problemset/problem/314/C

【题意】



让你从n个元素的数组中选出所有的不同的非递减子数列;

然后计算比这个子数列小的和它的长度一样长的数列的个数;

“小”的定义在题目里有说;

【题解】



设dp[i]表示以i作为非递减子数列的最后一个数的比它小的数列的个数;

则有递推式

dp[i] = (dp[1]+dp[2]+…+dp[i])*i+i;

写个树状数组,来快速求和就好;

要写出原数组,维护原数组;

不然求dp[i]比较麻烦;

最后输出∑dpi就好



【Number Of WA】



1(数组开小了)



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0),cin.tie(0)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1e6;
const int N1 = 1e5;
const int MOD = 1e9+7;

struct BIT{
    LL a[N+10];
    int lowbit(int x){
        return x&(-x);
    }

    void add(int x,LL y){
        while (x<=N){
            a[x] = (a[x]+y)%MOD;
            x+=lowbit(x);
        }
    }

    LL sum(int x){
        LL temp = 0;
        while (x>0){
            temp = (temp+a[x])%MOD;
            x-=lowbit(x);
        }
        return temp;
    }
}c;

int n;
LL dp[N+10];

int main(){
    //Open();
    Close();
    cin >> n;
    rep1(i,1,n){
        int x;
        cin >> x;
        LL temp = (c.sum(x)*x + x)%MOD;
        c.add(x,(temp-dp[x]+MOD)%MOD);
        dp[x] = temp;
    }
    cout << c.sum(N) << endl;
    return 0;
}